SAT (Fall 2023)
- Solving linear equations and linear inequalities — Basic example
- Solving linear equations and linear inequalities — Harder example
- Interpreting linear functions — Basic example
- Interpreting linear functions — Harder example
- Linear equation word problems — Basic example
- Linear equation word problems — Harder example
- Linear inequality word problems — Basic example
- Linear inequality word problems — Harder example
- Graphing linear equations — Basic example
- Graphing linear equations — Harder example
- Linear function word problems — Basic example
- Linear function word problems — Harder example
- Systems of linear inequalities word problems — Basic example
- Systems of linear inequalities word problems — Harder example
- Solving systems of linear equations — Basic example
- Solving systems of linear equations — Harder example
- Systems of linear equations word problems — Basic example
- Systems of linear equations word problems — Harder example
Watch Sal work through a harder Systems of linear equations word problem.
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- you can solve it with less time and effort. you can substract 4 children and 1 adult from each choice and see if the remaining children are double the adults since each of the remaining adults brought 2 children. think smarter not harder haha(52 votes)
- Agnes has 23 collectible stones, all of which are labradorite crystals or galena crystals. Labradorite crystals are worth $20 each, while galena crystals are worth $13 each. Agnes earns $439 by selling her entire collection. How many stones of each type did she sell?
I'm still stuck on this one problem....please help!(19 votes)
- 20 labradorite and 3 galena
because 20*20=400 and 3*13=39 so the total will be 439 which is exactly how much Agnes made(22 votes)
- Why are we given harder practice problems then what the example shows? This one looks more simple, but the other practice problems are so complicated sometimes. The SAT is coming soon, I just feel so under prepared..(17 votes)
- My solution:
Consider the number of adults to be x
Subsequently, the number of children can be written in terms of x as "2x + 2". (Each adult brings 2 children and one of the adults brought 2 extra children)
Now multiply the number of adults and children with their corresponding prices and equate that to 60.
That is "2(2x + 2) + 4x = 60"
Upon solving, you shall get x = 7 which implies that there are 7 adults. Thereafter, you can also find out the number of children which is "2x + 2" and that is 16. Thus, Option C is correct.(14 votes)
- Here's my solution:
*Note: a-1 = number of children who brought 2 children
subtract the one who brought 4(8 votes)
- Hi, is this problem you immediately knew that there will be only 1 adult with two kids so you did the math of 4+2x2=8.
I thought, however, that there will be two adults (parents) as my fist logical thing to think.
how could one understand that there is only one adult for each 2 kids?
- Don't overlook "the remaining adults brought 2 children EACH." So, it's children per adult regardless; singles or couples.(5 votes)
- [Instructor] Tickets for a play were $2 for each child and $4 for each adult. At one showing of the play, one adult brought four children and the remaining adults brought two children each. The total ticket sales from the children and adults was $60. How many children and adults attended the play? Alright, this is an interesting one. Okay, so let's just think about how much we've spent at the play and we know it has to add up to $60. Let's think about it in terms of the children and the adults and their admissions. So you have this one adult right over here that brought four children. So how much is that adult, how much is this family? Let's just assume it's a family. How much are they going to spend? Well, that one adult is going to spend $4 for their own ticket and then four children at $2 each. So plus four children times $2 per child, this is going to be $8 for the children's tickets plus $4 on theirs. They're going to spend $12. So that adult is going to spend $12. And then there's some remaining number of adults that brought two children. So let's just say r is I could say the remaining number of adults or the number of adults with two children. Adults with two children, that's r. So each of these adults with two children, how much are they going to spend? Well, they're each going to spend $4 on their own ticket for the adult and then they're gonna have two children at $2 each, so they're gonna spend $4 on the children's tickets. So they're gonna spend $8 in total. So each of these adults with two children is gonna spend $8 at the play and there's r of them. So they're going to spend $8 for each of these adults with two children and there are r of them. So this is the total amount of ticket sales from the adults with two children. And we add that to the ticket sales from this one adult with the four children and they're gonna have to add up to $60. So this is gonna have to add up to $60. Let's see, we can subtract 12 from both sides and so on the left, we'll be left with 8r is equal to 60 minus 12 is 48. Divide both sides by eight and you get r is equal to six. So we wanna be very careful, you might say, okay there was six adults these are just the adults with two children. There's six adults with two children, but there's another adult. There's another adult who brought four children. So there's a total of seven adults, seven adults total. Pardon my handwriting. Seven adults total. So we could just look at these choices, only one of these choices have seven adults. And we could verify that this would also amount to 16 children because this person up here, in magenta, they would bring four children, so you would have four children plus six adults brought two children, so six adults bringing two children each, that would amount to 12 children. And that indeed, that indeed does add up to be 16. And if you're under time pressure, you can see there is only one choice that has seven adults, so you could just pick that one.